Question

Find the flow rate Q when the flow velocity distribution u (m / s) in the pipeline with radius R (m) as shown in the figure below is expressed by the following equation as a function of the distance y (m) from the inner wall surface. However, U is the maximum flow velocity (m / s) in the central axis of the pipeline, and the flow velocity distribution u is symmetrical with respect to the central axis of the pipeline, and 0 ≦ y ≦ R. (Hint: The area of the gray part in the figure below (right) is dA and is represented by y and dy.)

When the inner diameter (= 2R) of the conduit is 101.6 mm and the flow rate Q = 10 L / min, find the maximum flow velocity U. (2 significant digits and unit)

U = U (2R – y)y R2 U i dyti u ro 1 day

Answer 1

U (2R-9) dA R² 2a (R-y) dy da = udA - 2aV/y(22-7)(y) by = (2xU [ 2R² y = 3Ry² + 43 dy y R d = { da R² 0 = 24V (R² y ² - Ry3 + 2 R² Q - TUR² 2 2R = 101-6mm R = 50.8 mm = 0.0508m Q = 10L/min = 10 x 10 -3 m3 1.67X10-4 m/s GOS U = 20 2 x 1.67810-4 0.041 m/s AR? 3-14% (0.0508)